Hessian is proportional to the metric everywhere

Question

Let $(\Omega^{n+1},g)$ be a compact Riemannian manifold with smooth boundary. Let $f\in C^{\infty}(\bar{\Omega})$ satisfies $\operatorname{Hess}f=\frac{1}{n+1}g.$ Suppose the minimum of $f$ occures at some interior point $x_0. $ I want to know why this equality implies that $$\operatorname{grad} f=\frac{1}{n+1}r\frac{\partial}{\partial r},$$ where $r$ is the distance to the point $x_0.$

Answer 1

For any point $x$ other than $x_0$, suppose $\gamma$ is the geodesic connecting $x_0$ and $x$ whose length is $r$. Let $V=\frac{\partial}{\partial r}$, and $w$ be any vector that is orthogonal to $v=V(x)$. Parallel transport $w$ along $\gamma$ to get a vector field (along $\gamma$) $W$.
Now we have $$\frac{d}{dr}(Wf)=VW(f)=\operatorname{Hess}(f)(V,W)$$ $$=\frac{1}{n+1}\langle V,W\rangle_g=0.$$ So $W(f)$ is a constant function along $\gamma$. However, we have $Wf(x_0)=df_{x_0}(W)=0$ as $x_0$ is a critical point. Thus, $W(f) \equiv 0$ along $\gamma$ which implies $\langle \operatorname{grad}(f),w\rangle_g=w(f)=W_x(f)=0$.
Similarly we have $$\frac{d}{dr}(Vf)=\frac{1}{n+1}.$$ Integrating along $\gamma$ we get $Vf(x)=\frac{r}{n+1}$ and $\langle \operatorname{grad}(f), v\rangle_g=v(f)=V_x(f)=\frac{r}{n+1}$. These results lead to the conclusion that $$\operatorname{grad}(f)=\frac{r}{n+1}V.$$